Dangers of local oscillator signal being stronger than radio frequency signal in mixer

Recently as an interview question for an RF engineer job, I was asked the following question.

What are the dangers of having a local oscillator power level much higher than that of the RF power level during mixing?

I did not feel as though I had a great answer to this question. The main concerns I could think of are as follows and are based on Poser's Microwave Circuits equation for the output of a single-ended Schottky diode mixer; (using the small signal approximation) (13.100)

$$\ i(t) = I_0 + G_d[v_{RF}(t)+v_{LO}(t) +\dfrac{G'_d}{2}[v_{RF}(t)+v_{LO}(t)]^2 + ... \$$

$$\ i(t) = I_0 + G_d[v_{RF}(t)+v_{LO}(t) + \dfrac{G'_d}{4}(V_{RF}^2(1+cos(2w_{RF}t))+V_{LO}^2(1+cos(2w_{LO}t)) + 2V_{RF}V_{LO}cos((w_{RF}-w_{LO})t) + 2V_{RF}V_{LO}cos((w_{RF}+w_{LO})t) \$$

I understand for a balanced mixer some of these terms could be removed/problems alleviated.

1. Coupling from LO to RF input could lead to strong radiation of LO signal
2. At the output of the mixer, there will be a term that is a copy of the LO signal. This term should be filtered out, but maybe if it's strong enough could still show up at the ADC
3. There may also be terms that are the square voltage of the LO signal and with either 0 frequency (DC) or double the frequency of the LO. These terms also should be filtered out, but if strong enough, perhaps these components could also still overpower the RF signal after filtering. In the case of the DC component, it could also bias the diodes/transistors used
4. In the case of unexpected components showing up in the signal, it could be possible for the ADC to be saturated/a loss in dynamic range

If anyone can think of additional possible answers or explain why mine are wrong, please do so.

• Did you start scribbling those equations in the job interview? (I'm trying to understand the relevance of you mentioning the job interview if any). In short, what was your job-interview reply (you have tagged it). Commented Jan 23, 2023 at 10:53

• Hi @Neil_UK thank you for the answer, if you don't mind me asking how were you able to use a $\dfrac{\lambda}{4}$ coupler in such a wide band filter? Commented Jan 25, 2023 at 12:52